Evaluate the Summation sum from x=1 to infinity of (1/3)^(x-1)

Solution:

Step 1:

To determine the sum of an infinite geometric series, we use the formula $S = \frac{a}{1 - r}$, where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio.

Step 2:

Identify the common ratio by using the formula $r = \frac{a_{x+1}}{a_x}$.

Step 2.1:

Insert the terms $a_x$ and $a_{x+1}$ into the formula to find $r$: $r = \frac{(\frac{1}{3})^{(x+1)-1}}{(\frac{1}{3})^{x-1}}$.

Step 2.2:

Proceed to simplify the expression.

Step 2.2.1:

Eliminate the shared base in the numerator and denominator.

Step 2.2.1.1:

Extract the common factor $(\frac{1}{3})^{x-1}$ from $(\frac{1}{3})^{(x+1)-1}$: $r = \frac{(\frac{1}{3})^{x-1} \cdot (\frac{1}{3})^{(x+1)-(x-1)}}{(\frac{1}{3})^{x-1}}$.

Step 2.2.1.2:

Remove the common factors.

Step 2.2.1.2.1:

Multiply by 1: $r = \frac{(\frac{1}{3})^{x-1} \cdot (\frac{1}{3})^{(x+1)-(x-1)}}{(\frac{1}{3})^{x-1} \cdot 1}$.

Step 2.2.1.2.2:

Cancel out the common factor: $r = \frac{\cancel{(\frac{1}{3})^{x-1}} \cdot (\frac{1}{3})^{(x+1)-(x-1)}}{\cancel{(\frac{1}{3})^{x-1}} \cdot 1}$.

Step 2.2.1.2.3:

Rewrite the simplified expression: $r = \frac{(\frac{1}{3})^{(x+1)-(x-1)}}{1}$.

Step 2.2.1.2.4:

Divide the remaining term by 1: $r = (\frac{1}{3})^{(x+1)-(x-1)}$.

Step 2.2.2:

Combine like terms: $r = (\frac{1}{3})^{(x-x+1)}$.

Step 2.2.3:

Simplify the exponent.

Step 2.2.3.1:

Apply the distributive property: $r = (\frac{1}{3})^{1}$.

Step 2.2.4:

The common ratio is thus $r = \frac{1}{3}$.

Step 3:

Since $|r| < 1$, the series is convergent.

Step 4:

Compute the first term of the series by substituting $x = 1$.

Step 4.1:

Substitute $x = 1$ into $(\frac{1}{3})^{x-1}$ to find $a$: $a = (\frac{1}{3})^{1-1}$.

Step 4.2:

Simplify the expression.

Step 4.2.1:

Subtract inside the exponent: $a = (\frac{1}{3})^{0}$.

Step 4.2.2:

Apply the exponent rule: $a = \frac{1^0}{3^0}$.

Step 4.2.3:

Any number raised to the power of 0 equals 1: $a = \frac{1}{3^0}$.

Step 4.2.4:

Simplify further: $a = \frac{1}{1}$.

Step 4.2.5:

The first term is $a = 1$.

Step 5:

Insert the values of $a$ and $r$ into the summation formula: $S = \frac{1}{1 - \frac{1}{3}}$.

Step 6:

Simplify the expression.

Step 6.1:

Simplify the denominator.

Step 6.1.1:

Express 1 as a fraction with a common denominator: $S = \frac{1}{\frac{3}{3} - \frac{1}{3}}$.

Step 6.1.2:

Combine the fractions over the common denominator: $S = \frac{1}{\frac{2}{3}}$.

Step 6.2:

Multiply the numerator by the reciprocal of the denominator: $S = 1 \cdot \frac{3}{2}$.

Step 6.3:

Final multiplication gives $S = \frac{3}{2}$.

Step 7:

The sum of the series can be expressed in various forms:

• Exact Form: $\frac{3}{2}$
• Decimal Form: $1.5$
• Mixed Number Form: $1 \frac{1}{2}$

Knowledge Notes:

To solve this problem, we need to understand the concept of an infinite geometric series and its convergence criteria. An infinite geometric series has the form $a + ar + ar^2 + ar^3 + \ldots$, where $a$ is the first term and $r$ is the common ratio between consecutive terms. The series converges if the absolute value of the common ratio $|r|$ is less than 1. The sum of a converging infinite geometric series can be calculated using the formula $S = \frac{a}{1 - r}$.

In this problem, we are given a series where each term is of the form $(\frac{1}{3})^{x-1}$. To find the common ratio $r$, we use the ratio of successive terms. We simplify the expression by canceling out common factors and applying exponent rules. Once we have determined that the series converges (because $|r| < 1$), we calculate the first term by substituting the initial value of $x$ into the term formula. Finally, we substitute the values of $a$ and $r$ into the sum formula and simplify to find the sum of the series.